Problem: Takahashi loves sorting.

He has a permutation (p_1,p_2,...,p_N) of the integers from 1 through N. Now, he will repeat the following operation until the permutation becomes (1,2,...,N):

* First, we will define high and low elements in the permutation, as follows. The i-th element in the permutation is high if the maximum element between the 1-st and i-th elements, inclusive, is the i-th element itself, and otherwise the i-th element is low.
* Then, let a_1,a_2,...,a_k be the values of the high elements, and b_1,b_2,...,b_{N-k} be the values of the low elements in the current permutation, in the order they appear in it.
* Lastly, rearrange the permutation into (b_1,b_2,...,b_{N-k},a_1,a_2,...,a_k).



How many operations are necessary until the permutation is sorted?

Constraints

* 1 ≤ N ≤ 2×10^5
* (p_1,p_2,...,p_N) is a permutation of the integers from 1 through N.

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 ... p_N


Output

Print the number of operations that are necessary until the permutation is sorted.

Examples

Input

5
3 5 1 2 4


Output

3


Input

5
5 4 3 2 1


Output

4


Input

10
2 10 5 7 3 6 4 9 8 1


Output

6

Solution (in CPP):

Solution:
#include <bits/stdc++.h>

#define pb(n) push_back(n)
#define maxn 200010

using namespace std;

vector<int> v;
int t[maxn],f[maxn],p[maxn],q[maxn];

int next(int x)
{
	return v[(x+1)%3];
}

int main()
{
	int n;
	scanf("%d", &n);
	
	for(int i=1; i<=n; ++i)
	{
		scanf("%d", &p[i]);
		q[p[i]]=i;
	}
	
	for(int i=n-1; i>0; --i)
	{
		if(t[i+1]==0)
		{
			if(q[i]>q[i+1])
			{
				t[i]=1;
				f[i]=i+1;
			}
		}
		else
			{
				bool czy=true;
				v.pb(q[f[i+1]]);
				v.pb(q[i]);
				v.pb(q[i+1]);

				sort(v.begin(),v.end());
				for(int j=0; j<3; ++j)
				{
					//cout << v[j] << endl;
					if(v[j]==q[f[i+1]])
					{
						if(next(j)!=q[i])
							czy=false;
					}
					if(v[j]==q[i])
					{
						if(next(j)!=q[i+1])
							czy=false;
					}
					if(v[j]==q[i+1])
					{
						if(next(j)!=q[f[i+1]])
							czy=false;
					}
				}
				v.clear();
				if(czy)
				{
					t[i]=t[i+1];
					f[i]=f[i+1];
				}
				else
				{
					t[i]=t[i+1]+1;
					f[i]=i+1;
				}
			}
		//	cout << t[i] << " " << f[i] << endl;
	}
	printf("%d\n", t[1]);
	return 0;
}